Tuning-Math Archive Section 1: 451

Message: 451 - Contents - Hide Contents Date: Thu, 28 Jun 2001 22:39:42 Subject: Re: questions about Graham's matrices From: Paul Erlich To: <tuning-math@xxxxxxxxxxx.xxx> Sent: Thursday, June 28, 2001 2:47 PM Subject: [tuning-math] Re: questions about Graham's matrices

> --- In tuning-math@y..., graham@m... wrote: >> monz wrote: >

>>> This is along the lines of what I was trying to get Paul to >>> understand a couple of different times in the past. It's not >>> necessary to always use prime-factors as the basis for lattice >>> metrics... any numbers that give even, consistent divisions >>> of the pitch-space *in SOME way* will do. The different ways >>> of dividing (and multiplying) produce different kinds of lattices. >

> Funny -- I would have said that this is something I was trying to get > Monz to understand, rather than something Monz was trying to get me > to understand . . . Maybe Monz could restate the context in which he > was trying to get me to understand this?

Oh... one time was back around when I asked you to show me how to prime-factorize my meantone formulae. As I said, I don't have the mathematical understanding to even imagine accurately, let alone describe, some of the vague latticing ideas I have. I'm trying. But I'm glad to see that somehow we managed to agree on this, without understanding each other! Cool. -monz Yahoo! GeoCities * [with cont.] (Wayb.) "All roads lead to n^0" _________________________________________________________ Do You Yahoo!? Get your free @yahoo.com address at Yahoo! Mail - The best web-based email! * [with cont.] (Wayb.)

Message: 452 - Contents - Hide Contents Date: Fri, 29 Jun 2001 05:45:37 Subject: Re: questions about Graham's matrices From: Paul Erlich --- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> But I'm glad to see that somehow we managed to agree on this, > without understanding each other! Cool. >

Yes, and I think it's a very important point. Rather than the basis being 3, 5, 7, it could just as easily be, say, 5/4, 6/5, 7/6, or any other basis that spans the 3D lattice (though consonant intervals are somewhat preferable here).

Message: 453 - Contents - Hide Contents Date: Fri, 29 Jun 2001 01:45:10 Subject: [tuning-math] agreeing without understanding (was: questions about Graham's matrices) From: monz

> From: Paul Erlich <paul@s...> > To: <tuning-math@xxxxxxxxxxx.xxx> > Sent: Thursday, June 28, 2001 10:45 PM > Subject: [tuning-math] Re: questions about Graham's matrices > > > --- In tuning-math@y..., "monz" <joemonz@y...> wrote: >

>> But I'm glad to see that somehow we managed to agree on this, >> without understanding each other! Cool. >>

> Yes, and I think it's a very important point.

I agree. We were both frustrated that we couldn't understand each other, in both cases, but it turns out that we were thinking basically the same thing all along.

> Rather than the basis being 3, 5, 7, it could just as > easily be, say, 5/4, 6/5, 7/6, or any other basis that > spans the 3D lattice (though consonant intervals are > somewhat preferable here).

Sure, these are all good. But I was thinking along the lines of, say, an axis representing a meantone generator, such as (3/2)/((81/80)^(1/4)) , for example. This is why, a few months back, I was so interested in Regener's work. It seems to be along these lines... transformations of ratio-space. -monz Yahoo! GeoCities * [with cont.] (Wayb.) "All roads lead to n^0" _________________________________________________________ Do You Yahoo!? Get your free @yahoo.com address at Yahoo! Mail - The best web-based email! * [with cont.] (Wayb.)

Message: 454 - Contents - Hide Contents Date: Fri, 29 Jun 2001 12:27 +0 Subject: Re: Hypothesis revisited From: graham@m... In-Reply-To: <9hgb48+f78j@e...> Dave Keenan wrote:

> I still don't understand this. So 10+41n includes ETs 10 51 61 71 ... > and 31+41n includes 31 72 113 ... Only the second looks anything like > Miracle to me.

They both cover this part of the scale tree 31 10 41 72 51 93 113 91 61 but branch differently at 41. So one is more closely associated with the Miracle family, but I don't think there's anything special about one unison vector as compared to the other. Probably I should ignore the generalisation, and take 31+10 and 10+31 as different ways of writing the same MOS.

> Graham replied:

>> There are two Canstas, 10+31n and 21+31n. >

> What could this mean when 31 and 72 aren't members of either of these > series? I'm very confused.

31 is the member where n=infinity. It enshrines the relationship 10+21=31, and hence this part of the scale tree 21 10 31 52 41 73 83 72 51 There are two different ways you can move on from Canasta, and different chromas suggest different branchings, but I've yet to see a deep reason for it. Graham

Message: 455 - Contents - Hide Contents Date: Fri, 29 Jun 2001 12:27 +0 Subject: Re: questions about Graham's matrices From: graham@m... In-Reply-To: <003101c1005d$50edba00$4448620c@a...> monz wrote:

> OK... how about "the *potential* generated scales" instead? > > But my point is... they're *approximately* 19, 22, and 41-equal, right?

They all contain MOS subsets with these numbers of notes.

> Huh? In a general sense, generators can be any interval, so > what does prime ordering have to do with it? Clarification on > this would be appreciated.

The period is a stand-in for the octave, and the generator is a stand-in for the twelfth (or fifth or whatever but not always). So it makes sense to put them in the same positions as the octave and fifth. In some common cases they really are an octave and fifth. Although you can define them the other way round, it's simpler not to.

>>>> | 19= | 22= | 41= >>>> --------------------------------- >>>> x | 0 | 1 | 1 >>>> y | 1 | 0 | 1 >>>> p | 0 | 1 | 1 >>>> q | 1 | 1 | 2

> I was hoping for another remedial-algebra lesson on how this > table could be calculated from the other data you gave. :)

An octave is 22x+19y steps. So when x=0 you have 19-equal and when y=0 you have 22-equal. When x=y you have 41-equal. As p is the same as x, it must have the same size in those temperaments. As q is the sum of x and y it'll either be 1 (from 1+0 or 0+1) or 2 (from 1+1). In algebraic terms (x) = (1)oct/22 or (0)oct/19 or (1)oct/41 (y) (0) (1) (1) (q) = (1 1)(x) = (1 1)(1)oct/22 = (1)oct/22 (p) (1 0)(y) (1 0)(0) (1) or (q) = (1 1)(0)oct/19 = (1)oct/19 (p) (1 0)(1) (0) or (q) = (1 1)(1)oct/41 = (2)oct/41 (p) (1 0)(1) (1) Graham

Message: 456 - Contents - Hide Contents Date: Fri, 29 Jun 2001 12:27 +0 Subject: Re: Hypothesis revisited From: graham@m... In-Reply-To: <9hg94q+1es0@e...> Paul wrote:

>> There are two Canstas, 10+31n and 21+31n. >

> Hmmm . . . what's the _real_ difference between these two?

How are you defining reality?

>> Are the FPBs different in this sense? >

> Yes -- look back a few days -- I showed that there was a schisma > difference between a few corresponding pitches in the two FPBs, even > though you're claiming the schisma as a chromatic unison vector > (hence one that isn't tempered out).

So does that mean the schisma isn't a valid chroma?

>> Yes, it would do. If you try tuning a 12-note meantone in cents > relative

>> to 12-equal, you'll see the pattern. >

> I see the pattern, but that doesn't make 1/12 octave the period of a > 12-note meantone . . . ?

It makes it the period of a linear temperament that includes 12-note meantone as a subset.

>>> Well you _should_ be able to find the generator without > specifying

>>> the chroma, but you need the chroma to select a particular MOS. >>

>> Indeed so! But the octave invariant matrix doesn't give you that >> particular MOS. >

> Sure it does! Just take the determinant (usually)! (Assuming you > already know the generator.)

Usually isn't good enough, we're looking for proof here. Besides, taking the determinant's cheating. It doesn't mean anything for octave-invariant matrices, but happens to be part of the result for octave-specific matrices.

>> the main result is the mapping in terms of >> generators. >

> Well, that does seem to be something very interesting you've found. > How can we get that without plugging in a chroma at all?

I'm hoping that always using a fifth for the top row will work. If not, framing the problem might help. We want to find a generator consistent with the simplest mapping, I suppose. Which means minimizing the determinant. We don't want it to go to zero, but that follows from the matrix being invertible. Is there any established theory of integer matrices, or discrete vector spaces, we can latch on to? Graham

Message: 457 - Contents - Hide Contents Date: Fri, 29 Jun 2001 15:06 +0 Subject: Vector update From: graham@m... Dividing through by the common factor of the whole normalised, inverted matrix does do the trick for my multiple-29 vectors. I think it's the *commatic* unison vectors that mean you have to do this. So my unison vector finder needs to be improved (surprise!) The second column of the normalised octave-specific inverse is always the same as the first column of the octave-invariant one, with an extra zero. I was forgetting the extra zero before. It may be this doesn't always work for really silly unison vectors, but it does for all the examples I've tried. The octave-specific column of the octave-specific matrix is important for getting the right scale-step mapping. This may be what was going wrong with the multiple-29 before. Whatever, it works now. I've got a rough and ready Excel spreadsheet showing this at <404 Not Found * [with cont.] Search for http://www.microtonal.co.uk/vectors.xls in Wayback Machine>. You need to install the Analysis ToolPack for the GCD function to work. Matrix operations work with the standard install. The Exchange Server at work is currently flaky, and although I have offline folders I don't seem to be able to get at them offline. So although I did read an e-mail from Monzo this morning, I can't reply to it. I think any commatic unison vector will do to get the generator mapping, so long as it's orthogonal to the other vectors. One good way of finding such is to try a 1 in every column until the determinant is non-zero. I'll try to include these changes in my Python code tonight. Python with the Numerical extensions is a good way of hacking this stuff, but the latter had disappeared from the FTP server last I checked, so I don't know how you'll get hold of them. The source code to MIDI Relay should include a matrix library for C++. Graham

Message: 458 - Contents - Hide Contents Date: Fri, 29 Jun 2001 19:25:19 Subject: Re: agreeing without understanding (was: questions about Graham's matrices) From: Paul Erlich --- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> > Sure, these are all good. But I was thinking along the > lines of, say, an axis representing a meantone generator, > such as (3/2)/((81/80)^(1/4)) , for example.

Well that, of course, would be the _right_ way to plot meantone tunings. Which is what I was trying to tell you while grudgingly figuring out for you how to plot meantone on the usual (3,5) axes using fractional exponents.

Message: 459 - Contents - Hide Contents Date: Fri, 29 Jun 2001 19:32:52 Subject: Re: Hypothesis revisited From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9hg94q+1es0@e...> > Paul wrote: >

>>> There are two Canstas, 10+31n and 21+31n. >>

>> Hmmm . . . what's the _real_ difference between these two? >

> How are you defining reality? Tuning. >

>>> Are the FPBs different in this sense? >>

>> Yes -- look back a few days -- I showed that there was a schisma >> difference between a few corresponding pitches in the two FPBs, even >> though you're claiming the schisma as a chromatic unison vector >> (hence one that isn't tempered out). >

> So does that mean the schisma isn't a valid chroma?

I'm not saying that . . . but first, can you determine which of the two scales (if either) is the "real" MIRACLE-41 (to within commatic unison vectors)?

>>> Yes, it would do. If you try tuning a 12-note meantone in cents >> relative

>>> to 12-equal, you'll see the pattern. >>

>> I see the pattern, but that doesn't make 1/12 octave the period of a >> 12-note meantone . . . ? >

> It makes it the period of a linear temperament that includes 12- note > meantone as a subset.

Oh -- but a very strange subset. Any "normal" subset should repeat exactly at the period . . . or that's how I've been thinking about this stuff.

>>>> Well you _should_ be able to find the generator without >> specifying

>>>> the chroma, but you need the chroma to select a particular MOS. >>>

>>> Indeed so! But the octave invariant matrix doesn't give you that >>> particular MOS. >>

>> Sure it does! Just take the determinant (usually)! (Assuming you >> already know the generator.) >

> Usually isn't good enough, we're looking for proof here. Besides, > taking the determinant's cheating. It doesn't mean anything for > octave-invariant matrices,

It doesn't mean anything?? It means a lot -- see the "Gentle Introduction" again . . .

> but happens to be part of the result for > octave-specific matrices. >

Part of the __________ result?

>>> the main result is the mapping in terms of >>> generators. >>

>> Well, that does seem to be something very interesting you've found. >> How can we get that without plugging in a chroma at all? >

> I'm hoping that always using a fifth for the top row will work. If not, > framing the problem might help. We want to find a generator consistent > with the simplest mapping, I suppose.

The simplest mapping? Not following you. The generator of an MOS is unique.

Message: 460 - Contents - Hide Contents Date: Fri, 29 Jun 2001 19:34:59 Subject: Re: Vector update From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:

> I think any commatic

You mean chromatic?

> unison vector will do to get the generator mapping, > so long as it's orthogonal to

You mean linearly independent from?

> the other vectors.

Message: 461 - Contents - Hide Contents Date: Fri, 29 Jun 2001 22:45 +0 Subject: Re: Vector update From: graham@m... Paul wrote:

>> I think any commatic >

> You mean chromatic? Yes.

>> unison vector will do to get the generator mapping, >> so long as it's orthogonal to >

> You mean linearly independent from?

I think so, but I didn't take good notes in that lecture.

>> the other vectors. > Graham

Message: 462 - Contents - Hide Contents Date: Fri, 29 Jun 2001 22:45 +0 Subject: Re: Hypothesis revisited From: graham@m... Paul wrote:

>>>> There are two Canstas, 10+31n and 21+31n. >>>

>>> Hmmm . . . what's the _real_ difference between these two? >>

>> How are you defining reality? > > Tuning.

There's no difference.

>> So does that mean the schisma isn't a valid chroma? >

> I'm not saying that . . . but first, can you determine which of the > two scales (if either) is the "real" MIRACLE-41 (to within commatic > unison vectors)?

Don't know, it's all on my Linux partition.

>> It makes it the period of a linear temperament that includes 12- > note

>> meantone as a subset. >

> Oh -- but a very strange subset. Any "normal" subset should repeat > exactly at the period . . . or that's how I've been thinking about > this stuff.

Yes, it takes 144 notes to get the 12 note scale. But it does prove the hypothesis that every set of vectors gives some linear temperament.

>> Usually isn't good enough, we're looking for proof here. Besides, >> taking the determinant's cheating. It doesn't mean anything for >> octave-invariant matrices, >

> It doesn't mean anything?? It means a lot -- see the "Gentle > Introduction" again . . .

I'll look it up.

>> but happens to be part of the result for >> octave-specific matrices. >>

> Part of the __________ result?

Octave-specific. It's the top left-hand corner.

>> I'm hoping that always using a fifth for the top row will work. If > not,

>> framing the problem might help. We want to find a generator > consistent

>> with the simplest mapping, I suppose. >

> The simplest mapping? Not following you. The generator of an MOS is > unique.

As long as it's unique, there's no problem. (Technically, it'll be +/-, but that's all negotiable) Graham

Message: 464 - Contents - Hide Contents Date: Fri, 29 Jun 2001 23:48:35 Subject: Re: Hypothesis revisited From: Dave Keenan --- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9hgb48+f78j@e...> > Dave Keenan wrote: >

>> I still don't understand this. So 10+41n includes ETs 10 51 61 71 ... >> and 31+41n includes 31 72 113 ... Only the second looks anything like >> Miracle to me. >

> They both cover this part of the scale tree > > 31 10 > > > 41 > > 72 51 > > 93 113 91 61

Ok. I can see that you are the one who is confused here. Miracle does not go outside of 31 41 72 93 113 Well Ok, it does go a tiny bit past 41, but nowhere near all the way to 10. Just as 5-EDO is nothing like a meantone. This is because, outside of 31 to 41(and-a-bit) there are better 7 or 11-limit approximations than the ones used by MIRACLE.

>> Graham replied:

>>> There are two Canstas, 10+31n and 21+31n. >>

>> What could this mean when 31 and 72 aren't members of either of these >> series? I'm very confused. >

> 31 is the member where n=infinity.

Huh? When n=oo 10+31n and 21+31n also go to oo. I think you must be talking your own language here. It enshrines the relationship

> 10+21=31, and hence this part of the scale tree > > > 21 10 > > > 31 > > 52 41 > 73 83 72 51 > > > There are two different ways you can move on from Canasta

Maybe so. But only one of them is MIRACLE.

Message: 466 - Contents - Hide Contents Date: Sat, 30 Jun 2001 06:37:16 Subject: Re: Hypothesis revisited From: Dave Keenan --- In tuning-math@y..., "Orphon Soul, Inc." <tuning@o...> wrote:

> On 6/29/01 7:48 PM, "Dave Keenan" <D.KEENAN@U...> wrote: > >>> 31 10 >>> >>> >>> 41 >>> >>> 72 51 >>>

>>> 93 113 91 61 >> >

> Shouldn't the bottom line be 103, 113, 92, 61?

Oh yes. Well spotted!

Message: 468 - Contents - Hide Contents Date: Sat, 30 Jun 2001 15:03:32 Subject: Re: Hypothesis revisited From: Graham Breed Dave Keenan wrote:

>> They both cover this part of the scale tree >> >> 31 10 >> >> >> 41 >> >> 72 51 >> >> 93 113 91 61 >

> Ok. I can see that you are the one who is confused here. Miracle does > not go outside of > > 31 > > > 41 > > 72 > > 93 113

I see a 41 there.

> Well Ok, it does go a tiny bit past 41, but nowhere near all the way > to 10. Just as 5-EDO is nothing like a meantone.

So decimal notation is now invalid? And blackjack isn't part of the family? 5-EDO may not be a meantone, but pentatonic scales certainly are.

> This is because, outside of 31 to 41(and-a-bit) there are better 7 > or 11-limit approximations than the ones used by MIRACLE. >>> Graham replied:

>>>> There are two Canstas, 10+31n and 21+31n. >>>

>>> What could this mean when 31 and 72 aren't members of either of > these

>>> series? I'm very confused. >>

>> 31 is the member where n=infinity. >

> Huh? When n=oo 10+31n and 21+31n also go to oo. I think you must be > talking your own language here.

Canasta is made up of 31 steps. For 10+31n, there are 10 of those at (n+1)/(10+31n) octaves and the other 21 are n/(10+31n) octaves. As n tends to infinity, both steps tend to 1/31 octaves.

> It enshrines the relationship

>> 10+21=31, and hence this part of the scale tree >> >> >> 21 10 >> >> >> 31 >> >> 52 41 >> 73 83 72 51 >> >> >> There are two different ways you can move on from Canasta >

> Maybe so. But only one of them is MIRACLE.

I've changed the way the temperaments are written to sweep all this under the carpet. Graham "I toss therefore I am" -- Sartre

Message: 469 - Contents - Hide Contents Date: Sat, 30 Jun 2001 23:29:39 Subject: Re: Hypothesis revisited From: Dave Keenan --- In tuning-math@y..., Graham Breed <graham@m...> wrote:

> So decimal notation is now invalid?

Of course not.

> And blackjack isn't part of the family?

Of course it is part of the family.

> 5-EDO may not be a meantone, but pentatonic scales certainly are.

Indeed, this is the crux of the confusion. The Stern-Brocot tree (considered as fractions of an octave) knows nothing about odd-limits (or any other kind), while the definition of Miracle, or meantone or any other temperament, must refer to them. The tree can tell us two different things about a temperament. (a) The number of notes in its MOS (b) The EDOs that are included in that temperament But we look up these things on the scale tree in two different ways. We need to know the range of generator sizes that are within the temperament. First we determine what limit we are using (say 7-odd for Miracle, 5-odd for meantone). Then we consider the maximum number of generators we are willing to chain to approximate these just intervals. (say 20 for Miracle and 11 for meantone). From this we can determine the range of generator sizes for which the temperament's mapping from primes to generators (Miracle [6, -7, -2] and meantone [1, 4]) gives us the best approximation. It is really the mapping from primes to generators that is the definition of the temperament. Once we have the two extreme generator sizes, we express these as fractions of an octave and mark them at the "bottom" of the tree (where the reals live). Draw straight lines up from these and the denominator of any fraction between those bounds gives us an ET within that temperament. The denominator of any fraction reachable by going up the tree from these, gives us the cardinality of a MOS in the temperament. So 10 and 11 and 21 are MOS cardinalities in Miracle temperament but certainly not EDO cardinalities. If that were the case, why stop at 10 and 11, why not go all the way back to 0 and 1? And remember that the SB tree has numerators and denominators. For convenience when talking about a particular temperament we drop the numerators. This might lead to confusion if we join together what are really disjoint parts of the tree, based on the denominators only.

> I've changed the way the temperaments are written to sweep all this under > the carpet. I'm glad.

-- Dave Keenan

Message: 470 - Contents - Hide Contents Date: Sat, 30 Jun 2001 18:39:50 Subject: Re: Hypothesis revisited From: monz I have a question for all of you mathematicians. I've just put up a Dictionary entry for LucyTuning. Definitions of tuning terms: LucyTuning, (c) 2... * [with cont.] (Wayb.) In it, I'd like to provide the calculation for the ratio of the LucyTuning "5th". Can this be simplified?: ( 2^(3 / (2*PI) ) ) * ( {2 / [2^(5 / (2*PI) ) ] } ^(1/2) ) -monz Yahoo! GeoCities * [with cont.] (Wayb.) "All roads lead to n^0" _________________________________________________________ Do You Yahoo!? Get your free @yahoo.com address at Yahoo! Mail - The best web-based email! * [with cont.] (Wayb.)

Message: 472 - Contents - Hide Contents Date: Sun, 01 Jul 2001 02:00:00 Subject: Re: Hypothesis revisited From: Paul Erlich --- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> I have a question for all of you mathematicians. > > I've just put up a Dictionary entry for LucyTuning. > Definitions of tuning terms: LucyTuning, (c) 2... * [with cont.] (Wayb.) > > > In it, I'd like to provide the calculation for the > ratio of the LucyTuning "5th". Can this be simplified?: > > ( 2^(3 / (2*PI) ) ) * ( {2 / [2^(5 / (2*PI) ) ] } ^(1/2) )

I think so. The LucyTuning "major third" is 2^(1/pi). Add two octaves to form the "major seventeenth": 2^(2+1/pi). Take the fourth root (since it's a meantone, the fifth will be the fourth root of the major seventeenth): 2^(1/2 + 1/(4*pi)). Is that right?

Message: 473 - Contents - Hide Contents Date: Sat, 30 Jun 2001 19:20:54 Subject: Re: Hypothesis revisited From: monz

> From: Paul Erlich <paul@s...> > To: <tuning-math@xxxxxxxxxxx.xxx> > Sent: Saturday, June 30, 2001 7:00 PM > Subject: [tuning-math] Re: Hypothesis revisited > > I think so. The LucyTuning "major third" is 2^(1/pi). > Add two octaves to form the "major seventeenth": 2^(2+1/pi). > Take the fourth root (since it's a meantone, the fifth > will be the fourth root of the major seventeenth): > 2^(1/2 + 1/(4*pi)). Is that right?

Thanks for this great explanation, Paul. Your answer is slightly different from the one Ed Borasky calculated with Derive: 2^( (2*pi) + 1 / (4*pi) ) -monz Yahoo! GeoCities * [with cont.] (Wayb.) "All roads lead to n^0" _________________________________________________________ Do You Yahoo!? Get your free @yahoo.com address at Yahoo! Mail - The best web-based email! * [with cont.] (Wayb.)